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Chapter 4: Problem 29

The forces on a small airplane (mass \(1160 \mathrm{kg}\) ) in horizontal flight heading eastward are as follows: gravity \(=16.000 \mathrm{kN}\) downward, lift \(=16.000 \mathrm{kN}\) upward, thrust \(=1.800 \mathrm{kN}\) eastward, and \(\mathrm{drag}=1.400 \mathrm{kN}\) westward. At \(t=0,\) the plane's speed is \(60.0 \mathrm{m} / \mathrm{s} .\) If the forces remain constant, how far does the plane travel in the next \(60.0 \mathrm{s} ?\)

Short Answer

Expert verified
Answer: The airplane travels approximately 4221 meters eastward in 60.0 seconds.

Step by step solution

01

Sum up horizontal forces to find net force

In this problem, we only need to analyze horizontal forces, which are thrust (eastward) and drag (westward). To find the net horizontal force (\(F_{net}\)), subtract the drag force from the thrust force: \(F_{net} = F_{thrust} - F_{drag} = 1800 - 1400 = 400\,\text{N}\).
02

Calculate the airplane's acceleration

Now that we have the net force, we can find the airplane's acceleration using Newton's second law, which states \(F = ma\). Rearrange the formula to solve for acceleration (\(a\)): \(a = \frac{F_{net}}{m} = \frac{400\,\text{N}}{1160\,\text{kg}} \approx 0.345\,\text{m/s}^2\)
03

Use kinematic equations to find the distance traveled

We are given the initial speed (\(v_0 = 60.0\,\text{m/s}\)), time (\(t = 60.0\,\text{s}\)), and acceleration (\(a = 0.345\,\text{m/s}^2\)). We can use the kinematic equation for the distance traveled, which is: $$ d = v_0t + \frac{1}{2}at^2 $$ Plug in the given values and compute the distance: $$ d = (60.0\,\text{m/s})(60.0\,\text{s}) + \frac{1}{2} (0.345\,\text{m/s}^2)(60.0\,\text{s})^2 \approx 3600\,\text{m}+ 621\,\text{m} = 4221\,\text{m} $$
04

Report the final answer

In the next 60.0 seconds, the airplane travels a distance of approximately 4221 meters eastward.

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Most popular questions from this chapter

In Fig. 4.15 an astronaut is playing shuffleboard on Earth. The puck has a mass of \(2.0 \mathrm{kg} .\) Between the board and puck the coefficient of static friction is 0.35 and of kinetic friction is \(0.25 .\) (a) If she pushes the puck with a force of \(5.0 \mathrm{N}\) in the forward direction, does the puck move? (b) As she is pushing, she trips and the force in the forward direction suddenly becomes \(7.5 \mathrm{N} .\) Does the puck move? (c) If so, what is the acceleration of the puck along the board if she maintains contact between puck and stick as she regains her footing while pushing steadily with a force of \(6.0 \mathrm{N}\) on the puck? (d) She carries her game to the Moon and again pushes a moving puck with a force of \(6.0 \mathrm{N}\) forward. Is the acceleration of the puck during contact more, the same, or less than on Earth? Explain. (tutorial: rough table)
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